Is Minimum Moves to Reach Target in Grid easy or hard?

The problem is classified as Hard because the brute-force BFS idea quickly fails for large inputs. The intended solution requires recognizing the mathematical pattern and converting the process into a reverse Euclidean reduction with logarithmic complexity.

Minimum Moves to Reach Target in Grid Python/Java solution

Most implementations follow the reverse reduction strategy. The algorithm repeatedly applies modulo operations to shrink the larger coordinate until it either reaches the start cell or becomes invalid. The logic is identical across Python, Java, C++, and Go, with only syntax differences.

How to solve Minimum Moves to Reach Target in Grid in O(log n)?

Work backward from the target coordinates instead of expanding from the start. If the allowed moves are additive, the previous state must be derived by subtracting the smaller coordinate from the larger one. Using modulo operations (tx %= ty or ty %= tx) compresses multiple reverse steps into one, producing an O(log n) solution.

What is the best approach for Minimum Moves to Reach Target in Grid?

The most efficient solution works backward from the target using a mathematical reduction similar to the Euclidean algorithm. Instead of simulating moves forward, repeatedly reduce the larger coordinate using modulo operations until it matches the start state. This approach runs in O(log(max(tx, ty))) time and uses constant space.

Is Minimum Moves to Reach Target in Grid asked at Google/Amazon/Meta?

Grid reachability and reverse-math problems appear frequently in interviews at companies like Google, Amazon, and Meta. Variants of additive coordinate transformations and Euclidean-style reductions are commonly used to test mathematical reasoning and optimization skills.

What data structure is used in Minimum Moves to Reach Target in Grid?

The brute-force interpretation uses a queue for BFS because the grid can be viewed as a graph of coordinate states. The optimal solution does not require complex data structures and instead relies on arithmetic operations and mathematical invariants.

What is the time complexity of Minimum Moves to Reach Target in Grid?

The optimal solution runs in O(log(max(tx, ty))) time because each step reduces one coordinate using a modulo operation, similar to the GCD algorithm. Space complexity is O(1) since only a few variables are tracked during the reverse reduction process.

Description Hints 3 Video 23:15

3609. Minimum Moves to Reach Target in Grid

Hard15.5% acceptance1 company

You are given four integers sx, sy, tx, and ty, representing two points (sx, sy) and (tx, ty) on an infinitely large 2D grid.

You start at (sx, sy).

At any point (x, y), define m = max(x, y). You can either:

Move to (x + m, y), or
Move to (x, y + m).

Return the minimum number of moves required to reach (tx, ty). If it is impossible to reach the target, return -1.

Example 1:

Input: sx = 1, sy = 2, tx = 5, ty = 4

Output: 2

Explanation:

The optimal path is:

Move 1: max(1, 2) = 2. Increase the y-coordinate by 2, moving from (1, 2) to (1, 2 + 2) = (1, 4).
Move 2: max(1, 4) = 4. Increase the x-coordinate by 4, moving from (1, 4) to (1 + 4, 4) = (5, 4).

Thus, the minimum number of moves to reach (5, 4) is 2.

Example 2:

Input: sx = 0, sy = 1, tx = 2, ty = 3

Output: 3

Explanation:

The optimal path is:

Move 1: max(0, 1) = 1. Increase the x-coordinate by 1, moving from (0, 1) to (0 + 1, 1) = (1, 1).
Move 2: max(1, 1) = 1. Increase the x-coordinate by 1, moving from (1, 1) to (1 + 1, 1) = (2, 1).
Move 3: max(2, 1) = 2. Increase the y-coordinate by 2, moving from (2, 1) to (2, 1 + 2) = (2, 3).

Thus, the minimum number of moves to reach (2, 3) is 3.

Example 3:

Input: sx = 1, sy = 1, tx = 2, ty = 2

Output: -1

Explanation:

It is impossible to reach (2, 2) from (1, 1) using the allowed moves. Thus, the answer is -1.

Constraints:

0 <= sx <= tx <= 10⁹
0 <= sy <= ty <= 10⁹

Asked at 1 company

Bloomberg

Topics

Math

Hints open one at a time

Hint 1

Work backwards from (tx, ty) to (sx, sy), undoing one move at each step.

Hint 2

If the larger coordinate >= 2 × (the smaller), undo by halving the larger; otherwise undo by subtracting the smaller from the larger.

Hint 3

Count these undo-steps until you hit (sx, sy) (return the count), or return -1 if you drop below or get stuck.

LanguageLoading editor…

1class Solution:
2    def minMoves(self, sx: int, sy: int, tx: int, ty: int) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

Input

Your code is saved on this device.